I Gauss' law seems to imply instantaneous electric field

[Bob44]

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire.

We wish to find an expression for the horizontal component of the electric field at a distance $\mathbf{r}$ from the sphere as it discharges.

By using the Lorenz gauge condition:
$$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$
we find the following retarded solutions to the Maxwell equations

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$ and $\partial_t\,\mathbf{j}(\mathbf{r},t)$, $\partial_t\mathbf{A}(\mathbf{r},t)$ are purely vertical then the horizontal component of the electric field is given by the retarded expression
$$\mathbf{E}(\mathbf{r},t)=-\nabla\phi=\frac{q(t-r/c)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{2}$$
Now let us compare eqn.$(2)$ with the result that we get from Gauss' law
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.\tag{3}$$
We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$
Eqn.$(2)$ is a causal retarded expression whereas eqn.$(4)$ is instantaneous.

Why do we have this discrepancy?

It seems that we have artificially obtained a causal solution $(2)$ by choosing the Lorenz gauge condition. The non-local expression $(4)$ obtained from Gauss' law did not assume any gauge condition.

 

[PeroK]

Have you assumed a uniform field across the surface of the sphere?

 

[Bob44]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

 

[PeroK]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

The discharge breaks the symmetry.

 

[TSny]

The potential at the field point $p$ at time $t$ is given by the retarded-time expression $\phi(r, t) = \large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$.

When evaluating $\nabla \phi$, we need to take into account that the numerator and denominator of $\large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$ depend on $r$.
$$\nabla \phi = \nabla \left[\frac{q(t-r/c)}{4\pi\epsilon_0 r}\right] = \frac{\nabla q(t-r/c)}{4\pi\epsilon_0 r} + \frac{ q(t-r/c)}{4\pi\epsilon_0 } \nabla \frac 1 r$$ Using the chain rule, you can show that $$\nabla q(t-r/c) = \dot q(t-r/c) \nabla(t-r/c) = -\frac 1 c \dot q(t-r/c) \hat{\mathbf{r}}$$ Here, $\dot q(t-r/c)$ is the rate at which $q$ is changing at the retarded time; that is, $\dot q(t-r/c) = -I(t-r/c)$.

So, we find, using $\nabla \large \frac 1 r = -\frac{\hat{\mathbf{r}}}{r^2}$, $$\nabla \phi =\frac{1}{4\pi\epsilon_0}\left[ \frac{I(t-r/c)}{rc} - \frac{q(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$

As a simple example, suppose the current is a constant, $I_0$, for all time. Then $q(t) = Q_0 - I_0t$ for all time t, where $Q_0$ is the charge at $t = 0$. Since the current is constant, the vector potential $\mathbf{A}$ is time independent. Thus $\partial_t \mathbf{A}$ does not contribute to the electric field. The charge at the retarded time can be written as $q(t-r/c) = Q_0 - I_0\cdot(t-r/c)$.

So, for this example, the electric field at point $p$ is $$\mathbf{E}_p = -\nabla \phi = -\frac{1}{4\pi\epsilon_0}\left[ \frac{I_0}{rc} - \frac{Q_0-I_0\cdot(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$ This simplifies to $$\mathbf{E}_p = \frac{Q_0-I_0t}{4\pi\epsilon_0r^2}\hat{\mathbf{r}} = \frac{q(t)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}$$ Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

 

[tech99]

So far as I can see, and I may be incorrect here, when the switch is closed there is oscillation, which is gradually damped out by resistive and radiation losses. The energy we extract by the oscillation was contained within a radius of half a wavelength of the oscillation.
If you discharge a sphere through a resistor, then it takes for ever to discharge. This is because you are now obtaining the energy returning from an infinite radius.
When you charge a sphere, the field will expand outwards for the whole time it is connected. When you discharge it, the field collapses for the time it is connected, returning the portion of the energy contained within a radius time x c. In some cases the time is fixed by oscillation in the wire and the time for a half cycle of oscillation.

 

[JimWhoKnew]

Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

A very nice demonstration.

In addition, and also following @PeroK's remark in #4, we can see that by considering a large "virtual" sphere of radius $R$ around the origin,$$\int\mathbf{E}\cdot d\mathbf{A}$$ on it will remain constant at the time between the the closing of the switch (assumed to be located near the charge) and the arrival of this information to the point where the wire exits the sphere (no charge goes in or out of the sphere during this interval). Therefore Equation (4) in OP is evidently incorrect during this interval. Since$$I(t-R/c)=0$$during this interval, equation (2) of OP is valid and $~\mathbf{E}(t,R)~$ is the same as before, as expected by causality.